1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
181
182
183
184
185
186
187
188
189
190
191
192
193
194
195
196
197
198
199
200
|
{
"metadata": {
"name": ""
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 1: Bonding in Solids"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 1.1, Page number 1.4"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from math import pi\n",
"\n",
"#Variable declartion\n",
"E_a = 502 #first ionization energy of A(kJ/mol)\n",
"E_b = -335 #electron affinity for B atom(kJ/mol)\n",
"d = 3*10**-10 #inter-ionic seperation(m)\n",
"E = 8.85*10**-12 #permittivity of free space(C/N -m)\n",
"e = 1.602*10**-19#charge of an electron(C)\n",
"\n",
"#Calculations\n",
"C_e = (-e**2)/(4*pi*E*d) #Coulumbic attraction energy(J/ion pair)\n",
"C_e_kj = C_e *6.022*10**23*10**-3 #Converting C_e to KJ/mol\n",
"n_e = E_a + E_b + C_e_kj #Net change in energy per mole\n",
"\n",
"#Results\n",
"print \"Net change in energy is =\",round(n_e,3),\"kJ/mol\"\n",
"print \"As the net change in energy is negative, the molecule will be stable\"\n",
"print \"\\nThe solution differs because of rounding-off of the digits in the textbook\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Net change in energy is = -296.224 kJ/mol\n",
"As the net change in energy is negative, the molecule will be stable\n",
"\n",
"The solution differs because of rounding-off of the digits in the textbook\n"
]
}
],
"prompt_number": 19
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 1.2, Page number 1.4"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from math import pi \n",
"\n",
"#Variable declaration\n",
"IP_K = 4.1 #ionization energy of K(eV)\n",
"EA_Cl = 3.6 #electron affinity of Cl(eV)\n",
"e = 1.602*10**-19 #charge of an electron(C)\n",
"Eo = 8.85*10**-12 #permittivity of free space(C/N -m)\n",
"\n",
"#Calcuations\n",
"del_E = IP_K - EA_Cl\n",
"#Ec = 0.5 = -e/4*pi*Eo*R\n",
"R = e/(4*pi*Eo*del_E)\n",
"\n",
"#Results\n",
"print \"(a) If their total energy is to be zero, then del_E = Ec\"\n",
"print \"(b) The seperation between the ion pair is\", round((R/1E-9),2), \"nm\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(a) If their total energy is to be zero, then del_E = Ec\n",
"(b) The seperation between the ion pair is 2.88 nm\n"
]
}
],
"prompt_number": 7
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 1.3, Page number 1.5"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from math import pi\n",
"\n",
"#Variable declaration\n",
"ro = 2.36*10**-10 #inter ionic equilibrium distance(m)\n",
"IP = 5.14 #ionization energy of Na(eV)\n",
"EA = 3.65 #electron affinity of Cl(eV)\n",
"e = 1.602*10**-19 #charge of an electron(C)\n",
"Eo = 8.85*10**-12 #permittivity of free space(C/N -m)\n",
"\n",
"#Calculations\n",
"Ue = (-e**2)/(4*pi*Eo*ro*e) #Potential energy in eV\n",
"Be = -Ue - IP + EA\n",
"\n",
"#Result\n",
"print \"Bond energy for NaCl molecule is\",round(Be,2),\"eV\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Bond energy for NaCl molecule is 4.61 eV\n"
]
}
],
"prompt_number": 9
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 1.4, Page number 1.18"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from math import pi\n",
"\n",
"#Variable declaration\n",
"ro = 0.281*10**-9 #equilibrium seperation between the ion pair(m)\n",
"IP = 5.14 #ionization energy of Na(eV)\n",
"EA = 3.61 #electron affinity of Cl(eV)\n",
"A = 1.748 #Madelung constant\n",
"n = 9. #born replusive exponent\n",
"e = 1.602*10**-19 #charge of an electron(C)\n",
"Eo = 8.85*10**-12 #permittivity of free space(C/N -m)\n",
"\n",
"#Calculations\n",
"Ce = (A*(e**2)*(1-(1/n)))/(4*pi*Eo*ro) #joules\n",
"Ce_ev = Ce/e #Converting joules to eV\n",
"\n",
"#Result\n",
"print \"The cohesive energy of NaCl molecule is\",round(Ce_ev,2),\"eV\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The cohesive energy of NaCl molecule is 7.97 eV\n"
]
}
],
"prompt_number": 22
}
],
"metadata": {}
}
]
}
|